please provide information?

Question

a^(3)+b^(3)+c^(3)=d^(3)
like 3^(3)+4^(3)+5^(3)=6^(3)
please provide a website link or general formula
to find out the numbers(natural) in the above equation.

Answer 1

Every positive rational number is the sum of 3 positive rational cubes, so given an integer d, you can at least express d^3 as the sum of 3 positive rational cubes. The general solution of your equation is only known if you allow for rational solutions, and then you can search through those to find integral ones. First, we should observe that finding solutions to your equation is the same as finding solutions to a^3 + b^3 = e^3 + f^3, since then we can write c=-e and d=f and find a solution to your equation. The equation a^3 + b^3 = e^3 + f^3 obviously has the trivial solutions a=b=0, e=-f and also a=e, b=f, and others of this sort. Apart from these trivial solutions, the other solutions are all given by

a = x (1 - (y-3z)(y^2+3z^2)), b=x((y+3z)(y^2+3z^2)-1), e=x((y+3z)-(y^2+3z^2)^2), and f=x((y^2+3z^2)^2-(y-3z)),

where x, y, and z are any rational numbers. Plugging in y=z=1, x=1/3, we find your solution:

3^3+5^3=(-4)^3+6^3.

Ramanujan gave a partial solution to your equation a^3+b^3+c^3=d^3:

a=3x^2+5xy-5y^2, b=4x^2-4xy+6y^2, c=5x^2-5xy-3y^2, d=6x^2-4xy+4y^2. Plutting in a=2, b=1, we get the solution
17^3+14^3+7^3=20^3

These questions you keep posting are from an area of number theory called Diophantine Equations. You should check out a book on them sometime.

Answer 2

Man,

U can get it using XL. Jus a lil bit of effort

Answer 3

TRY PURPLEMATH.COM FOR YOUR ALGEBRA NEEDS.